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The following problem arises in molecular biology. The surface of a bacterium consists of several sites at which foreign molecules-some acceptable and some not-become attached. We consider a particular site and assume that molecules arrive at the site according to a Poisson process with parameter \(\lambda\). Among these molecules a proportion \(\alpha\) is acceptable. Unacceptable molecules stay at the site for a length of time that is exponentially distributed with parameter \(\mu_{1}\), whereas an acceptable molecule remains at the site for an exponential time with rate \(\mu_{2}\). An arriving molecule will become attached only if the site is free of other molecules. What percentage of time is the site occupied with an acceptable (unacceptable) molecule?

Short Answer

Expert verified
The percentage of time the site is occupied by an acceptable molecule is given by: Percentage of time occupied by acceptable molecule = \(100 \times \frac{位伪\times \frac{1}{渭_2}}{位 \times ( \frac{伪}{渭_2} + \frac{1-伪}{渭_1})}\) And the percentage of time the site is occupied by an unacceptable molecule is given by: Percentage of time occupied by unacceptable molecule = \(100 \times \frac{位(1-伪)\times \frac{1}{渭_1}}{位 \times ( \frac{伪}{渭_2} + \frac{1-伪}{渭_1})}\) Use these formulas with given values of 位, 伪, 渭鈧, and 渭鈧 to calculate the percentages.

Step by step solution

01

Identifying the given parameters

We are given the following parameters: 1. Molecules arrive according to a Poisson process with parameter 位. 2. A proportion 伪 of the arriving molecules is acceptable. 3. Unacceptable molecules stay for an exponentially distributed time with parameter 渭鈧. 4. Acceptable molecules stay for an exponentially distributed time with parameter 渭鈧.
02

Calculate probabilities related to arrivals

Given the proportion 伪 of acceptable molecules, we can say that the probability of a molecule being acceptable upon arrival is P(A) = 伪. Therefore, the probability of a molecule being unacceptable is P(U) = 1 - 伪.
03

Calculate probabilities related to departure

Since the length of time an unacceptable molecule stays attached is exponentially distributed with parameter 渭鈧, we can say that the departure rate of an unacceptable molecule is 位_U = 位(1 - 伪)渭鈧. Similarly, since the length of time an acceptable molecule stays attached is exponentially distributed with parameter 渭鈧, we can say that the departure rate of an acceptable molecule is 位_A = 位伪渭鈧.
04

Calculate probabilities of site occupation

Using the provided information, we can now calculate the probabilities of the site being occupied by an acceptable molecule and an unacceptable molecule. In steady state, we can write: P(Occupied by acceptable molecule) = \(\frac{\text{Arrival rate of acceptable molecules} \times \text{average time occupied by acceptable molecule}}{\text{Total Arrival rate} \times (\text{average time occupied by acceptable molecule} + \text{average time occupied by unacceptable molecule})}\) P(Occupied by acceptable molecule) = \(\frac{位伪\times \frac{1}{渭_2}}{位 \times ( \frac{伪}{渭_2} + \frac{1-伪}{渭_1})}\) Similarly, P(Occupied by unacceptable molecule) = \(\frac{位(1-伪)\times \frac{1}{渭_1}}{位 \times ( \frac{伪}{渭_2} + \frac{1-伪}{渭_1})}\)
05

Calculate the percentages of time the site is occupied

Now that we have probabilities for site occupation by an acceptable and unacceptable molecule, we can calculate the percentage of time the site is occupied by each type: Percentage of time occupied by acceptable molecule = \(100 \times P(\text{Occupied by acceptable molecule})\) Percentage of time occupied by unacceptable molecule = \(100 \times P(\text{Occupied by unacceptable molecule})\) Using values of 位, 伪, 渭鈧, and 渭鈧, these percentages can be calculated to determine the percentage of time the site is occupied by an acceptable and unacceptable molecule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is a key concept in understanding the duration that molecules remain attached at a site. In this context, the exponential distribution describes the time until an event occurs, such as a molecule detaching from a bacterium's surface. It's characterized by a rate parameter, usually denoted as \( \mu \). This rate is essentially the average number of events (detachments in this case) that happen per unit time.

Here's a breakdown of why the exponential distribution is important in our problem:
  • **Memoryless Property:** This means the probability of a molecule detaching in the next moment is the same, regardless of how long it has already been attached. This is crucial for stochastic models, like those involving random arrivals and departures.
  • **Parameter \( \mu \):** It influences how long molecules will typically stay attached. A higher \( \mu \) means a shorter average attachment time.
In our scenario, unacceptable molecules have a detachment rate of \( \mu_1 \), whereas acceptable ones have \( \mu_2 \). This distinction affects how we calculate the chances of these molecules being at the attachment site at any given time.
Molecular Biology
In molecular biology, understanding interactions at the molecular level is crucial. The problem at hand reflects how molecules attach and detach from a bacterium's surface, a fundamental process in cellular interactions. Here, we examine a specific site on a bacterium which can accept or reject incoming molecules.

Consider the following aspects:
  • **Foreign Molecule Attachment:** Molecules arrive and attempt to attach based on their nature. Acceptable and unacceptable types differ, impacting the biological functionality of the surface.
  • **Acceptance Probability:** Not all molecules are equal; each has a probability \( \alpha \) of being acceptable. This is crucial for biological filtering processes.
These processes aren't just academic; they're foundational to how bacteria interact with their environments. Through this problem, we understand how different molecules impact the biological equilibrium of bacterial surfaces.
Stochastic Processes
Stochastic processes are mathematical objects usually defined as a set of random variables. In this problem, the arrival and attachment of molecules can be modeled using a Poisson process, a classic example of a stochastic process.

Here's why stochastic processes are essential:
  • **Poisson Process:** This specific type of stochastic process describes events happening independently at a constant average rate, \( \lambda \). In molecular biology, it models the random nature of molecules arriving at a surface.
  • **Randomness and Predictability:** While each event is random, the process itself has a predictable average behavior. This allows us to calculate probabilities of occupancy at the site.
By understanding stochastic processes, we can make sense of seemingly random molecular arrivals and predict how long the bacterium's site will be occupied by different molecule types. These insights are applied in modeling biological systems and designing experiments.

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Most popular questions from this chapter

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

Consider a set of \(n\) machines and a single repair facility to service these machines. Suppose that when machine \(i, i=1, \ldots, n\), fails it requires an exponentially distributed amount of work with rate \(\mu_{i}\) to repair it. The repair facility divides its efforts equally among all failed machines in the sense that whenever there are \(k\) failed machines each one receives work at a rate of \(1 / k\) per unit time. If there are a total of \(r\) working machines, including machine \(i\), then \(i\) fails at an instantaneous rate \(\lambda_{i} / r\) (a) Define an appropriate state space so as to be able to analyze the preceding system as a continuous-time Markov chain. (b) Give the instantaneous transition rates (that is, give the \(\left.q_{i j}\right)\). (c) Write the time reversibility equations. (d) Find the limiting probabilities and show that the process is time reversible.

Consider a system of \(n\) components such that the working times of component \(i, i=1, \ldots, n\), are exponentially distributed with rate \(\lambda_{i} .\) When a component fails, however, the repair rate of component \(i\) depends on how many other components are down. Specifically, suppose that the instantaneous repair rate of component \(i, i=1, \ldots, n\), when there are a total of \(k\) failed components, is \(\alpha^{k} \mu_{i}\) (a) Explain how we can analyze the preceding as a continuous-time Markov chain. Define the states and give the parameters of the chain. (b) Show that, in steady state, the chain is time reversible and compute the limiting probabilities.

Consider two \(\mathrm{M} / \mathrm{M} / 1\) queues with respective parameters \(\lambda_{i}, \mu_{i}, i=1,2 .\) Suppose they share a common waiting room that can hold at most three customers. That is, whenever an arrival finds her server busy and three customers in the waiting room, she goes away. Find the limiting probability that there will be \(n\) queue 1 customers and \(m\) queue 2 customers in the system. Hint: Use the results of Exercise 28 together with the concept of truncation.

Suppose that a one-celled organism can be in one of two states-either \(A\) or \(B\). An individual in state \(A\) will change to state \(B\) at an exponential rate \(\alpha ;\) an individual in state \(B\) divides into two new individuals of type \(A\) at an exponential rate \(\beta .\) Define an appropriate continuous- time Markov chain for a population of such organisms and determine the appropriate parameters for this model.

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